Answer
a. 1.17 M $LiCl$
b. 0.123 M $C_6H_{12}O_6$
c. $4.53 \times 10^{-3}M = 0.00453M$ $NaCl$
Work Step by Step
a.
$C(mol/L) = \frac{n(moles)}{volume (L)} = \frac{3.25 moles}{2.78L} = 1.17 $
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b.
1. Calculate the molar mass $(C_6H_{12}O_6)$:
12.01* 6 + 1.008* 12 + 16.00* 6 = 180.16g/mol
2. Calculate the number of moles $(C_6H_{12}O_6)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 28.33}{ 180.16}$
$n(moles) = 0.1572$
3. Find the concentration in mol/L $(C_6H_{12}O_6)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.1572}{ 1.28} $
$C(mol/L) = 0.123$
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c.
$122.4$ mL = $122.4 \times 10^{-3}$ L = 0.1224 L
1. Calculate the molar mass $(NaCl)$:
22.99* 1 + 35.45* 1 = 58.44g/mol
2. Calculate the number of moles $(NaCl)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.0324}{ 58.44}$
$n(moles) = 5.544\times 10^{- 4}$
3. Find the concentration in mol/L $(NaCl)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 5.544\times 10^{- 4}}{ 0.1224} $
$C(mol/L) = 4.53\times 10^{- 3}$
