Question #7b152

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Question #7b152

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Answer 1 · 2017-02-06T17:03:21

In the ORS ratio of

$salt : glucose = 1 : 12$ $" salt ":"glucose"=1:12$

So the percentage of salt in the mixture
$= \frac{1}{13} \times 100 %$ $=1/13xx100%$

In the salt ( $N a C l$ $NaCl$ ) the ratio of $(C l) : (N a) = 35.5 : 23$ $(Cl):(Na)=35.5:23$
So the fraction of (Cl) in the salt is $= \frac{35.5}{35.6 + 23} = \frac{35.5}{58.5}$ $=35.5/(35.6+23)=35.5/58.5$

So the percentage of $C l$ $Cl$ in the mixture
$= \frac{35.5}{58.5} \times \frac{1}{13} \times 100 % \approx 4.66 %$ $=35.5/58.5xx1/13xx100%~~4.66%$

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