to prepare 500cm^3 of glucose solution of concerntration 1M. The chemical formula of glucose is C6H12O6.calculate (1)the relative molecular mass and molar mass.(2)the amount of glucose which will be in the solution.(3)the mass of glucose to be weighed
To calculate the answers, we'll need some information:
1) The relative molecular mass (Molar Mass) of glucose.
2) The formula mass of glucose.
3) The volume and concentration of the solution.
First, let's calculate the relative molecular mass and molar mass of glucose (C6H12O6).
1) The relative molecular mass of glucose:
The relative molecular mass is the sum of the atomic masses of all the atoms in the compound.
Relative molecular mass of C6H12O6 = (6 * atomic mass of carbon) + (12 * atomic mass of hydrogen) + (6 * atomic mass of oxygen)
Looking up the atomic masses:
Atomic mass of carbon (C) = 12.01 g/mol
Atomic mass of hydrogen (H) = 1.01 g/mol
Atomic mass of oxygen (O) = 16.00 g/mol
Using these values, we can now calculate the relative molecular mass of glucose.
Relative molecular mass of glucose = (6 * 12.01) + (12 * 1.01) + (6 * 16.00)
= 72.06 + 12.12 + 96.00
= 180.18 g/mol
2) The molar mass of glucose:
The molar mass is the mass of one mole of a substance and is given by the same value as the relative molecular mass.
Molar mass of glucose = 180.18 g/mol
Now, let's move on to the next part:
3) The amount of glucose in the solution:
To prepare a 500 cm^3 (or 500 mL) solution with a concentration of 1 M, we need to calculate the amount of glucose required.
Amount of glucose (in moles) = Concentration (in mol/L) * Volume (in L)
= 1 mol/L * 0.5 L
= 0.5 mol
Finally, let's calculate the mass of glucose to be weighed:
Mass of glucose = Amount of glucose (in moles) * Molar mass of glucose
= 0.5 mol * 180.18 g/mol
= 90.09 g (rounded to two decimal places)
Therefore, to prepare a 500 cm^3 glucose solution with a concentration of 1 M, you will need to weigh 90.09 grams of glucose.
To answer your questions, let's go step by step:
(1) The relative molecular mass (RMM) is the sum of the atomic masses of all the atoms in a molecule. To calculate the RMM of glucose (C6H12O6), we need to find the atomic masses of carbon (C), hydrogen (H), and oxygen (O) and multiply them by their respective subscripts. The atomic masses are:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.008 g/mol
- Oxygen (O): 16.00 g/mol
Calculating the RMM:
6 x (C: 12.01 g/mol) + 12 x (H: 1.008 g/mol) + 6 x (O: 16.00 g/mol) = 180.18 g/mol
So, the relative molecular mass (RMM) of glucose is 180.18 g/mol.
The molar mass of a substance is the mass of one mole of that substance. In this case, the molar mass of glucose is the same as its RMM, which is 180.18 g/mol.
(2) Now, let's calculate the amount of glucose (in moles) that will be in the solution.
Concentration is defined as the amount of solute (glucose, in this case) per unit volume of solution. The given concentration is 1 M, which means 1 mole of glucose is present in 1 liter of solution.
We want to prepare 500 cm^3 (cubic centimeters) of the glucose solution. To convert cm^3 to liters, we need to divide by 1000, since there are 1000 cm^3 in 1 liter.
So, the volume in liters is: 500 cm^3 ÷ 1000 cm^3/L = 0.5 L
Since the concentration is 1 M, for every 1 liter of solution, we have 1 mole of glucose. Therefore, in 0.5 L of solution, we will have 0.5 moles of glucose.
(3) Finally, let's calculate the mass of glucose to be weighed.
Since we know the molar mass of glucose is 180.18 g/mol, and we want to prepare 0.5 moles of glucose, we can multiply these two values to find the mass:
0.5 moles x 180.18 g/mol = 90.09 grams
So, the mass of glucose to be weighed is 90.09 grams.