need help with these problems:

The concentration of glucose in blood is usually about 0.10% (w/v). If 7.0 L of blood are in an average man, how many grams of glucose are in his blood?

To obtain the solute for 118 g of a 19.2% (w/w) solution of sugar we would weigh out ________ g of sugar.

How many moles of sodium chloride are in 993 mL of a 2.88 M solution of sodium chloride?

A salt solution is evaporated to dryness. If 79.4 g of the solution leaves 1,015 mg of dry salt, what is the percent concentration (w/w) of the salt in the solution?

0.1% w/v is 0.1 g/100 mL.

(0.1/100 mL) x 7000 mL = ?g glucose.

(grams solute/grams soln) x 100 = % w/w
(g solute/118) x 100 = 19.2
Solve for g solute.

M = mols/L by defnition.
Substitute M and L (convert 993 mL to L) and solve for moles.

Sure! Let's go through each problem step-by-step.

1. The concentration of glucose in blood is given as 0.10% (w/v). This means that in every 100 mL of blood, there are 0.10 grams of glucose.
To find the number of grams of glucose in 7.0 L of blood, we need to convert 7.0 L to mL.
1 L = 1000 mL, so 7.0 L = 7.0 x 1000 mL = 7000 mL.
Now we can calculate the number of grams of glucose in 7.0 L of blood:
0.10 g/100 mL x 7000 mL = 7.0 g
Therefore, there are 7.0 grams of glucose in his blood.

2. In a 19.2% (w/w) solution of sugar, the concentration means that 19.2 grams of sugar are present in every 100 grams of the solution.
You have 118 grams of the solution, and you want to find out how much sugar is present in it.
Let's set up a proportion:
19.2 g sugar/100 g solution = x g sugar/118 g solution
Cross-multiply and solve for x:
(19.2 g)(118 g solution) = (100 g)(x g sugar)
2265.6 g = 100x
x = 2265.6 g / 100
x = 22.66 g
Therefore, to obtain the solute for 118 g of a 19.2% (w/w) solution of sugar, you would weigh out 22.66 grams of sugar.

3. In a 2.88 M solution of sodium chloride, the concentration means that there are 2.88 moles of sodium chloride in every 1 liter of solution.
You have 993 mL of the solution and you want to find out how many moles of sodium chloride are present in it.
First, convert 993 mL to liters:
993 mL = 993/1000 L = 0.993 L
Now we can calculate the number of moles of sodium chloride:
2.88 mol/L x 0.993 L = 2.86 moles
Therefore, there are 2.86 moles of sodium chloride in 993 mL of the 2.88 M solution.

4. To find the percent concentration (w/w) of the salt in the solution, we need to calculate the ratio of the mass of the salt to the mass of the solution.
We are given that 79.4 g of the solution leaves 1,015 mg (which is equivalent to 1.015 g) of dry salt.
Let's set up a proportion:
(79.4 g solution)/(1,015 g salt) = (x g solution)/100 g salt
Cross-multiply and solve for x:
(79.4 g)(100 g salt) = (1,015 g salt)(x g solution)
7940 g = 1015x
x = 7940 g / 1015
x = 7.8 g
Therefore, the percent concentration (w/w) of the salt in the solution is 7.8%.

Sure, I can help you with these problems. Let's go through each one step by step.

1. The concentration of glucose in blood is given as 0.10% (w/v). This means that for every 100 mL (or 0.1 L) of blood, there are 0.10 grams of glucose. We are given that there are 7.0 L of blood in an average man, so we can set up a proportion to find the number of grams of glucose in his blood.

0.10 grams of glucose x grams of glucose
--------------------- = -------------------
0.1 liters 7.0 liters

Cross-multiplying, we get:

0.10 * 7.0 = 0.1x

Simplifying:

0.70 = 0.1x

Dividing both sides by 0.1:

x = 0.70 / 0.1 = 7.0 grams

Therefore, there are 7.0 grams of glucose in the man's blood.

2. To obtain the solute for a 19.2% (w/w) solution of sugar, we need to weigh out a certain amount of sugar. This means that for every 100 grams of the solution, there are 19.2 grams of sugar.

Let's assume we need a total of 118 grams of the solution. To find the weight of sugar needed, we can set up a proportion:

19.2 grams of sugar x grams of sugar
------------------- = ----------------
100 grams 118 grams

Cross-multiplying, we get:

19.2 * 118 = 100x

Simplifying:

2265.6 = 100x

Dividing both sides by 100:

x = 2265.6 / 100 = 22.656 grams

Therefore, we would weigh out approximately 22.656 grams of sugar.

3. For this problem, we are given the concentration of a sodium chloride solution as 2.88 M, which means there are 2.88 moles of sodium chloride in 1 liter of the solution. We are given 993 mL of the solution, so we need to calculate the number of moles of sodium chloride in that volume.

To do this, we first convert the volume from milliliters to liters:

993 mL = 0.993 L

Now we can set up a proportion:

2.88 moles of sodium chloride x moles of sodium chloride
----------------------------- = ---------------------------
1 liter 0.993 liters

Cross-multiplying, we get:

2.88 * 0.993 = x

Simplifying:

2.85684 ≈ x

Therefore, there are approximately 2.85684 moles of sodium chloride in 993 mL of the solution.

4. In this problem, a salt solution is evaporated to dryness, leaving behind dry salt. We are given that the mass of the solution is 79.4 grams and the mass of the dry salt is 1,015 mg (or 1.015 grams). We need to find the percent concentration (w/w) of the salt in the solution.

To do this, we divide the mass of the dry salt by the mass of the solution and multiply by 100:

Percent concentration (w/w) = (mass of dry salt / mass of solution) * 100

Plugging in the given values:

Percent concentration (w/w) = (1.015 grams / 79.4 grams) * 100

Calculating:

Percent concentration (w/w) = 1.28%

Therefore, the percent concentration (w/w) of the salt in the solution is 1.28%.