A sodium nitrate solution is 21.5% (by mass) of nano3 (molar mass = 85.00 g/mol) and the solution has a density of 1.08 g/ml. calculate the molarity (m) of the solution.

Question

Ezequiel

Chemistry

28 May, 19:51

A sodium nitrate solution is 21.5% (by mass) of nano3 (molar mass = 85.00 g/mol) and the solution has a density of 1.08 g/ml. calculate the molarity (m) of the solution.

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Shaniya Jensen · Answer 1

Answer is: molarity of sodium nitrate solution is 2.73 M.

V (NaNO₃) = 1 L · 1000 mL/L = 1000 mL.

d (NaNO₃) = 1.08 g/mL.

ω (NaNO₃) = 21.5% : 100% = 0.215.

mr (NaNO₃) = V (NaNO₃) · d (NaNO₃).

mr (NaNO₃) = 1000 mL · 1.08 g/mL.

mr (NaNO₃) = 1080 g.

m (NaNO₃) = ω (NaNO₃) · mr (NaNO₃).

m (NaNO₃) = 0.215 · 1080 g.

m (NaNO₃) = 232.2 g.

n (NaNO₃) = m (NaNO₃) : M (NaNO₃).

n (NaNO₃) = 232.2 g : 85.00 g/mol.

n (NaNO₃) = 2.73 mol.

c (NaNO₃) = n (NaNO₃) : V (NaNO₃).

c (NaNO₃) = 2.73 mol : 1 L.

c (NaNO₃) = 2.73 mol/L.