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Question #7dec0

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Aug 25, 2017

$n = 0.93 moles$ $n = 0.93 "moles"$

Explanation:

Recall $\to M o l a r i t y = \frac{no of moles}{volume of solution}$ $-> Molarity = "no of moles"/"volume of solution"$

$\Rightarrow M = \frac{n}{v}$ $rArr M = n/v$

Where $M = 1.55 M or 1.55 m o l L^{- 1}$ $M = 1.55M or 1.55 molL^-1$

$\times \times x n = ?$ $color(white)(xxxxx) n = ?$

$\times \times x v = 600 M L = \frac{600 M L}{1000 M L} \times 1 L = 0.6 L$ $color(white)(xxxxx) v = 600ML = (600ML)/(1000ML) xx 1L = 0.6L$

$\Rightarrow M = \frac{n}{v}$ $rArr M = n/v$

Making 'n' the subject

$\Rightarrow n = M \times v$ $rArr n = M xx v$

$\Rightarrow n = 1.55 m o l L^{- 1} \times 0.6 L$ $rArr n = 1.55 molL^-1 xx 0.6L$

$rArr n = 1.55 mol cancel(L^-1) xx 0.6cancelL$

$rArr n = (1.55 xx 0.6) mol$

$n = 0.93 "moles"$

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