Question #11d6b

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Question #11d6b

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Answer 1 · 2014-12-08T20:47:47

The answer is $296.3 g$ $296.3g$ .

Saline solution is a mixture of table salt ( $N a C l$ $NaCl$ ) and water ( $H_{2} O$ $H_2O$ ).
The solution's percent concentration by mass, defined as the mass of solute (in this case, $N a C l$ $NaCl$ ) divided by the total mass of the solution and multiplied by 100%, is said to be $1.25 %$ $1.25%$ .

The total mass of the solution is equal to

$m_{s o l u t i o n} = m_{N a C l} + m_{w a t e r}$ $m_(solut ion) = m_(NaCl) + m_(water)$

Knowing that $c % = \frac{m_{s o l u t e}}{m_{s o l u t i o n}} \cdot 100 %$ $c% = m_(solute)/(m_(solut ion)) * 100%$ , and that the

mass of the solution is $300 g$ $300g$ , we can determine the mass of $N a C l$ $NaCl$ by

$m_{N a C l} = \frac{c \cdot m_{s o l u t i o n}}{100} = \frac{1.25 \cdot 300}{100} = 3.75 g r a m s$ $m_(NaCl) = (c * m_(solut ion))/100 = (1.25 * 300)/100 = 3.75grams$

This means that the mass of water is equal to

$m_{w a t e r} = m_{s o l u t i o n} - m_{N a C l} = 300 - 3.75 = 296.3 g r a m s$ $m_(water) = m_(solut ion) -m_(NaCl) = 300- 3.75 = 296.3 grams$ , which makes sense since the bulk of this particular solution must be represented by water, given the fact that its percent concentration by mass is $1.25 %$ $1.25%$ .

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Question #11d6b

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