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Calculate the molarity of each solution.(a) 22.6 g of C12H22O11 in 0.442 L of
Chapter 13, Problem 61P(choose chapter or problem)
Calculate the molarity of each solution.
(a) 22.6 g of \(\mathrm {C_{12}H_{22}O_{11}}\) in 0.442 L of solution
(b) 42.6 g of NaCl in 1.58 L of solution
(c) 315 mg of \(\mathrm {C_6H_{12}O_6}\) in 58.2 mL of solution
Questions & Answers
QUESTION:
Calculate the molarity of each solution.
(a) 22.6 g of \(\mathrm {C_{12}H_{22}O_{11}}\) in 0.442 L of solution
(b) 42.6 g of NaCl in 1.58 L of solution
(c) 315 mg of \(\mathrm {C_6H_{12}O_6}\) in 58.2 mL of solution
ANSWER:Step 1 of 4
Known:
a) \(\mathrm{m}\left(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}\right)=22.6\mathrm{\ g}\mathrm{\ V}(\text{ solution })=0.442\mathrm{\ L}\)
b) \(\mathrm{m}(\mathrm{NaCl})=42.6\mathrm{\ g}\mathrm{\ V}(\text{ solution })=1.58\mathrm{\ L}\)
c) \(\mathrm{m}\left(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\right)=315\mathrm{\ mg\ }\mathrm{V}(\text{ solution })=58.2\mathrm{\ mL}\)
Calculus:
\(c / \text { molarity }=\frac{\mathrm{n} / \text { moles solute }}{\mathrm{V} / \text { volumes solution }}\)
We will also need the standard formula for moles:
\(\mathrm{n} / \text { moles solute }=\frac{\mathrm{m} / \text { mass }}{\mathrm{M} / \mathrm{molar} \text { mass }}\)