Answer
a. 0.062 M $LiNO_3$
b. 0.675 M $C_2H_6O$
c. $6.898\times 10^{- 4}M$ $KI$
Work Step by Step
a.
$C(mol/L) = \frac{n(moles)}{volume (L)} = \frac{0.38 moles}{6.14L} = 0.062M $
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b.
1. Calculate the molar mass $(C_2H_6O)$:
12.01* 2 + 1.008* 6 + 16* 1 = 46.07g/mol
2. Calculate the number of moles $(C_2H_6O)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 72.8}{ 46.07}$
$n(moles) = 1.58$
3. Find the concentration in mol/L $(C_2H_6O)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 1.58}{ 2.34} $
$C(mol/L) = 0.675$
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c.
$112.4$ mL = $112.4 \times 10^{-3}$ L = 0.1124 L
1. Calculate the molar mass $(KI)$:
39.1* 1 + 126.9* 1 = 166.0g/mol
2. Calculate the number of moles $(KI)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.01287}{ 166.0}$
$n(moles) = 7.753\times 10^{- 5}$
3. Find the concentration in mol/L $(KI)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 7.753\times 10^{- 5}}{ 0.1124} $
$C(mol/L) = 6.898\times 10^{- 4}M$
