The solubility of a salt, \text{NaCl}, at 298\text{ K} is 6.0\text{ mol dm}^{-3}. What mas… — JAMB UTME

The solubility of a salt, $\text{NaCl}$ , at $298\text{ K}$ is $6.0\text{ mol dm}^{-3}$ . What mass of $\text{NaCl}$ must be dissolved in $250\text{ cm}^3$ of water to form a saturated solution at this temperature? $[\text{Molar mass of NaCl} = 58.5\text{ g mol}^{-1}]$

A
$1.50\text{ g}$
$87.75\text{ g}$ Answer
C
$351.00\text{ g}$
D
$14.63\text{ g}$

Answer

The mass of

\text{NaCl}

required to prepare a saturated solution is

87.75\text{ g}

A saturated solution contains the maximum mass of solute dissolved at a specified temperature. At

298\text{ K}

1.0\text{ dm}^3

of saturated solution requires

6.0\text{ mol}

\text{NaCl}

. For

250\text{ cm}^3

(

0.25\text{ dm}^3

), the required amount is

1.5\text{ mol}

. Multiplying

1.5\text{ mol}

58.5\text{ g mol}^{-1}

yields

87.75\text{ g}

Step-by-Step Solution

Convert the volume of water from

\text{cm}^3

\text{dm}^3

\text{Volume} = \frac{250\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.25\text{ dm}^3

Solubility is expressed per

\text{dm}^3

, so the volume must be in

\text{dm}^3

Calculate the amount of

\text{NaCl}

in moles required to saturate

0.25\text{ dm}^3

of water.

\text{Moles} = 6.0\text{ mol dm}^{-3} \times 0.25\text{ dm}^3 = 1.5\text{ mol}

The number of moles is obtained by multiplying molar concentration by volume in

\text{dm}^3

Convert moles of

\text{NaCl}

to mass using its molar mass.

\text{Mass} = 1.5\text{ mol} \times 58.5\text{ g mol}^{-1} = 87.75\text{ g}

Mass is calculated by multiplying the amount in moles by the molar mass.

Key Concept

Calculating solute mass required for saturation using solubility in

\text{mol dm}^{-3}

and molar mass.

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Answer

Step-by-Step Solution

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