Calculate the molarity of 1.5% w/v solution of PbCl2?

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Calculate the molarity of 1.5% w/v solution of PbCl2?

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Answer 1 · 2017-04-28T07:59:23

We ASSUME that a $100 \cdot m L$ $100*mL$ volume of SOLUTION contains a $1.5 \cdot g$ $1.5*g$ mass of solute..........

Explanation:

And thus $molarity$ $"molarity"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$

$= \frac{\frac{1.5 \cdot g}{278.10 \cdot g \cdot m o l^{- 1}}}{0.100 \cdot L^{- 1}} = 5.4 \times 10^{- 2} \cdot m o l \cdot L^{- 1}$ $=((1.5*g)/(278.10*g*mol^-1))/(0.100*L^-1)=5.4xx10^-2*mol*L^-1$ .

Lead chlorides are pretty insoluble. A solution of this concentration may represent a saturated solution.

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Calculate the molarity of 1.5% w/v solution of PbCl2?

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