# Question 7b152

Feb 6, 2017

In the ORS ratio of

$\text{ salt ":"glucose} = 1 : 12$

So the percentage of salt in the mixture
=1/13xx100%

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

So the percentage of $C l$ in the mixture
=35.5/58.5xx1/13xx100%~~4.66%#