We’re being asked to **calculate the molarity (M)** of a 7.92% (wt/wt) ethylene glycol (C_{2}H_{6}O_{2}) solution

Recall that ** molarity** is the ratio of the moles of solute and the volume of solution (in liters). In other words:

$\overline{){\mathbf{molarity}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{L}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

We’re given the mass percent of C_{2}H_{6}O_{2} in water, **7**** .92%**. Recall that

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{mass}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}{\mathbf{O}}_{\mathbf{2}}}{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}{\mathbf{x}}{\mathbf{100}}}$

This means in **100 g of solution**, we have **7.92 g C _{2}H_{6}O_{2} **.

**The steps we need to do for this solution are:**

**Step 1***:* Calculate the moles of C_{2}H_{6}O_{2}

**Step 2***:* Calculate the volume of the solution (in L).

**Step 3***:* Calculate the molarity of the solution.

A 7.92 mass % aqueous solution of ethylene glycol (HOCH_{2}CH_{2}OH) has a density of 1.15 g/mL. Calculate the molarity of the solution. Give your answer to 2 decimal places.

a. 0.36 m

b. 0.42 m

c. 1.92 m

d. 1.47 m

e. 1.21 m

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