Question

If all the water in 430.0 mL of a 0.45 M NaCI solution evaporates what is the mass of NaCI will remain

Answer 1

Answer:

11.31 g.

Explanation:

Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.

M = (no. of moles of solute)/(V of the solution (L)).

∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).

∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass) = (0.45 M)(0.43 L)(58.44 g/mol) = 11.31 g.

Answer 2

Hello!

If all the water in 430.0 mL of a 0.45 M NaCI solution evaporates what is the mass of NaCI will remain

We have the following data:  

M (Molarity) = 0.45 M (or 0.45 mol/L)

m1 (mass of the solute) = ? (in grams)

V (solution volume) = 430.0 mL → 0.43 L

MM (molar mass of NaCl) = 23u + 35.44u = 58.44u (or 58.44 g/mol)

Now, let's apply the data to the formula of Molarity, let's see:

$M = \dfrac{m_1}{MM*V}$

$0.45\:mol/L = \dfrac{m_1}{58.44\:g/mol*0.43\:L}$

$m_1 = 0.45\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}/\diagup\!\!\!\!L*58.44\:g/mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*0.43\:\diagup\!\!\!\!L$

$m_1 = 11.30814 \to \boxed{\boxed{m_1 \approx 11.31\:g}}\Longleftarrow(mass\:of\:the\:solute)\:\:\:\:\:\:\bf\green{\checkmark}$

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Answer:

The mass of NaCl is approximately 11.31 grams

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