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} $
$C(mol/L) = 0.716$
