Question

How much dry solute would you take to prepare each of the following solutions from the dry solute and the solvent?

-127mL of 0.100M NaNO3

-129g of 0.220m NaNO3

-129g of 0.220m NaNO3

-How much solvent would you take to prepare the solution in part B?

-How much solvent would you take to prepare the solution in part C?

Answer 1

A. 127 ml of 0.100M NaNO₃

M = (Weight in gram / Mol. wt.) (1000 / vol. in ml)

0.1 = (W / 85) (1000 / 127)

W = (0.1 x 85 x 127) / 1000 = 1.0795 gram

B and C. 129g of 0.220m NaNO₃ - Both of the questions are appearing similar. The symbol used for denoting molarity is 'M'. 'm' was used for denoting molality but it is obsolete now, insttead of it, mol/Kg is used. Anyways, I am going to solve the question in both the ways (assuming 0.220m as 0.220 molar and molal both).

1. If it is 0.220 molar NaNO₃

M = (129 / 85) (1000 / V)

0.22 = (129 x 1000) / (85 x V)

V = 129000 / (85 x 0.22) = 6898.4 ml = approx. 7 litre

2. If it is 0.220 molal NaNO₃

Molality = (No. of moles) / wt. of solvent in Kg = (wt. / Mol. wt.) / wt. of solvent in Kg

0.220 = (129 / 85) / W

W = 129 / 85 x 0.22 = 6.8984 Kg solvent