How much KCl in grams is in 0.337 L of a 5.80 % mass percent KCl solution? (Assume that the density of the solution is 1.05 g/mL.)

Question

Accepted Answer

GIVEN: 5.80% KCl by mass
0.337 L solution

[latex] d =\frac{1.05 g}{mL}[/latex]

FIND: g KCl

RELATIONSHIPS USED

[latex]\frac{5.80 g\,\mathrm{KCI}}{100\,\mathrm{g}\,\mathrm{solution}}[/latex] (given mass percent, written as a fraction)

[latex] \frac{105 g}{mL} [/latex] (given density)

1 mL = [latex]10^{-3}[/latex] L (Table 2.2)

[latex] 0.337 \cancel{\mathrm{L\; solution}} imes \frac{1 \cancel{mL}}{10^{-3} \cancel{L}} imes \frac{1.05 \cancel{g}}{ \cancel{mL}} imes \frac{5.80 g KCl}{ 100 \cancel{\mathrm{g \; solution}}} \ =20.5 g KCl [/latex]

The units (g KCl) are correct. The magnitude of the answer makes sense because 0.337 L is a bit more than 300 g of solution. Each 100 g of solution contains about 6 g KCl, Therefore the answer should be a bit more than 18 g.

TABLE 2.2 SI Prefix Multipliers
Prefix	Symbol	Multiplier
tera-	T	1,000,000,000,000	$(10^{12})$
giga-	G	1,000,000,000	$(10^9)$
mega-	M	1,000,000	$(10^6)$
kilo-	k	1,000	$(10^3)$
deci-	d	0.1	$(10^{-1})$
centi-	c	0.01	$(10^{-2})$
milli-	m	0.001	$(10^{-3})$
micro-	$\mu$	0.000001	$(10^{-6})$
nano-	n	0.000000001	$(10^{-9})$
pico-	p	0.000000000001	$(10^{-12})$
femto-	f	0.000000000000001	$(10^{-15})$

Question 13.12: How much KCl in grams is in 0.337 L of a 5.80 % mass percent......

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