Question

calculate the molality of 2.89g of NACl dissolved in .159L of H2O​

Answer 1

Answer:

Explanation:

0.159 L of h2o - 2.89 g of NaCl

1mL of h2o - 2.89/159 g of NaCl

= 0.018 g of NaCl

Moles of NaCl = 0.018/58.5 = 0.00031 mol

We know density of water is 1 g/mL

Mass of water 159 g

m = 0.00031/159

= 2×10^-6 ( approx)