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