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A 0.176 mol sample of Kr gas is contained in a 8.00 L flask at room temperature and pressure. What is the density of the gas, in grams/liter, under these conditions?

1 Answer

1s2s2p · Surya K.

Feb 23, 2018

$ρ = 1.84 g l L^{- 1}$ $rho=1.84gcolor(white)(l)L^-1$

Explanation:

First, we need to find the mass of $Kr$ $"Kr"$ by using the equation:

$\frac{m}{M_{r}} = n$ $m/M_r=n$ , where:

$m$ $m$ = mass ( $g$ $g$ )
$M_{r}$ $M_r$ = molar mass ( $g l m o l^{- 1}$ $gcolor(white)(l)mol^-1$ )
$n$ $n$ = number of moles ( $m o l$ $mol$ )

$m = n M_{r}$ $m=nM_r$

$m (Kr) = n (Kr) M_{r} (Kr) = 0.176 \cdot 83.8 = 14.7488 g$ $m("Kr")=n("Kr")M_r("Kr")=0.176*83.8=14.7488g$

$ρ = \frac{m}{V} = \frac{14.7488}{8} = 1.8436 \approx 1.84 g l L^{- 1}$ $rho=m/V=14.7488/8=1.8436~~1.84gcolor(white)(l)L^-1$

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