Question #198683

A student calculated the molarity of a solution prepared by dissolving 0.730 mol of table sugar (sucrose, C_{12}H_{22}O_{11}) in 1.8x10^3 mL of water as 4.06x10^-4 M C_{12}H_{22}O_{11}.

Expert's answer

"V=1.8 \\times 10^3 \\;mL = 1.8 \\;L \\\\\n\nn = 0.730 \\;mol"

Proportion:

0.730 mol – 1.8 L

x mol – 1.0 L

"x = \\frac{0.730 \\times 1.0}{1.8}=0.4055\\;M"

Student’s answer is wrong.

The right answer is 0.4055 M.

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