## Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company

# Chapter 4 - Questions and Problems - Page 179: 4.64

#### Answer

(a) 1.37 M (b) 0.426 M (c) 0.716 M

#### Work Step by Step

(a) $150$ mL = $150 \times 10^{-3}$ L = 0.15 L 1. Calculate the molar mass $(CH_3OH)$: 12.01* 1 + 1.008* 3 + 16* 1 + 1.008* 1 = 32.042g/mol 2. Calculate the number of moles $(CH_3OH)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 6.57}{ 32.042}$ $n(moles) = 0.205$ 3. Find the concentration in mol/L $(CH_3OH)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.205}{ 0.15}$ $C(mol/L) = 1.37$ (b) $220$ mL = $220 \times 10^{-3}$ L = 0.22 L 4. Calculate the molar mass $(CaCl_2)$: 40.08* 1 + 35.45* 2 = 110.98g/mol 5. Calculate the number of moles $(CaCl_2)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 10.4}{ 110.98}$ $n(moles) = 0.0937$ 6. Find the concentration in mol/L $(CaCl_2)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.0937}{ 0.22}$ $C(mol/L) = 0.426$ (c) $85.2$ mL = $85.2 \times 10^{-3}$ L = 0.0852 L 7. Calculate the molar mass $(C_{10}H_8)$: 12.01* 10 + 1.008* 8 = 128.164g/mol 8. Calculate the number of moles $(C_{10}H_8)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 7.82}{ 128.164}$ $n(moles) = 0.061$ 9. Find the concentration in mol/L $(C_{10}H_8)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.061}{ 0.0852}$ $C(mol/L) = 0.716$

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