Answer
a. 2.50 M glucose solution.
b. 1.00 M HCl solution.
c. 2.14 M NaOH solution.
Work Step by Step
a.
- Write the molarity expression and evaluate it:
$$Molarity(M) = \frac{moles \space of \space solute}{liters \space of \space solution}$$
$$= \frac{0.500 \space mole \space glucose}{ 0.200 \space L \space solution} = 2.50 \space M \space glucose \space solution$$
b.
- Calculate or find the molar mass for $ HCl $:
$ HCl $ : ( 35.45 $\times$ 1 )+ ( 1.008 $\times$ 1 )= 36.46 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 73.0 \space g \times \frac{1 \space mole}{ 36.46 \space g} = 2.00 \space moles$$
- Write the molarity expression and evaluate it:
$$Molarity(M) = \frac{moles \space of \space solute}{liters \space of \space solution}$$
$$= \frac{2.00 \space moles \space HCl}{ 2.00 \space L \space solution} = 1.00 \space M \space HCl \space solution$$
c.
- Calculate or find the molar mass for $ NaOH $:
$ NaOH $ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 40.00 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 30.0 \space g \times \frac{1 \space mole}{ 40.00 \space g} = 0.750 \space mole$$
$1 \space L = 1000 \space mL$
$$Volume(L) = 350. \space mL \times \frac{1 \space L}{1000 \space mL} = 0.350 \space L$$
- Write the molarity expression and evaluate it:
$$Molarity(M) = \frac{moles \space of \space solute}{liters \space of \space solution}$$
$$= \frac{0.750 \space mole \space NaOH}{ 0.350 \space L \space solution} = 2.14 \space M \space NaOH \space solution$$