Q4.you will be asked to prepare 25.0 mL of a 0.019 M solution of sodium chloride during this experiment....

a.calculate mass of sodium chloride needed to prepare solution.

(0.25L)(0.019M NaCl/L)*(58.5g NaCl/1mole NaCl)=0.278 g NaCl....is that right?

someone just replied it was 2.3 but how do i get that?

0.025 L x 0.019 M x 58.442 = 0.02776 which rounds to 0.028 g to two significant figures (2 s.f. in 0.019 and in 0.025).

given a 4.00stock of sodiam chloride NACL, how would you preparw 100ml of 1.oom

To calculate the mass of sodium chloride (NaCl) needed to prepare the solution, you can use the formula:

Mass = Volume (in liters) x Concentration (in moles/liter) x Molar mass of NaCl.

In this case, the volume is given as 25.0 mL, which needs to be converted to liters:

1 L = 1000 mL

So, 25.0 mL = 25.0 mL x (1 L / 1000 mL) = 0.025 L

Now you can substitute the values into the formula:

Mass = 0.025 L x 0.019 moles/L x 58.5 g/mole = 0.028 moles x 58.5 g/mole = 1.6435 g

Therefore, the correct mass of sodium chloride needed to prepare the solution is approximately 1.6435 grams.

To calculate the mass of sodium chloride needed to prepare the solution, you can use the formula:

Mass = (Volume) x (Concentration) x (Molar mass)

Given:
Volume = 25.0 mL = 0.025 L (since 1 mL = 0.001 L)
Concentration = 0.019 M
Molar mass of NaCl = 58.5 g/mol

So, using the formula:

Mass = (0.025 L) x (0.019 M) x (58.5 g/mol)
Mass = 0.027825 g

Therefore, the correct answer is approximately 0.028 g NaCl, not 2.3 g.