Question #959f8

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Question #959f8

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Answer 1 · 2016-03-15T10:58:04

$m a s s = 2.18 g C a C l_{2} \cdot 6 H_{2} O$ $mass=2.18 g CaCl_2*6H_2O$

Explanation:

Take the molar mass of $C a C L_{2} \cdot 6 H_{2} O$ $CaCL_2*6H_2O$
$C a = 1$ $Ca = 1$ x $40 = 40$ $40=40$
$C l = 2$ $Cl=2$ x $35 = 70$ $35=70$
$H_{2} O = 6$ $H_2O=6$ x $18 = 108$ $18=108$
Total= $218$ $218$ grams per mole
2. All other data needed to solve the problem are given:
$V = 100 m l = 0.100 L i$ $V=100 ml=0.100 Li$
$M = 0.1 \frac{m o l}{L i}$ $M=0.1 (mol)/(Li)$
3. Plug in data using the formula:
$M = \frac{n}{L i s o l u t i o n}$ $M=color(red)n/(Li solution)$
but:
$n = \frac{m a s s}{m o l a r m a s s}$ $color(red)n=(mass)/(molar mass)$
4. Therefore; plug in the value to derive new formula to find the mass of $C a C l_{2} \cdot 6 H_{2} O$ $CaCl_2*6H_2O$
$M = \frac{\frac{m a s s}{m o l a r m a s s}}{L i s o l u t i o n}$ $M=((mass)/(molar mass))/(Li solution)$
5. Rearrange the formula isolating $m a s s$ $mass$
$m a s s = m o l a r m a s s$ $mass=molar mass$ x $M$ $M$ x $L i s o l u t i o n$ $Li solution$
6. Plug in values and ensure to cancel out units to obtain $m a s s = 2.18 g C a C l_{2} \cdot 6 H_{2} O$ $mass=2.18 g CaCl_2*6H_2O$

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Question #959f8

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