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Question
question 9
what is the concentration (in m and mass %) of the resulting 96 ml solution when you add 5.0 grams of naoh to 95 ml of water? show your work and include all units in calculations. you may use the math editor. (density of water = 1.0 g/ml).
Step1: Calculate the moles of NaOH
The molar mass of NaOH ($M$) is $M = 23+16 + 1=40\space g/mol$.
Using the formula $n=\frac{m}{M}$, where $m = 5.0\space g$ (mass of NaOH) and $M = 40\space g/mol$.
$n=\frac{5.0\space g}{40\space g/mol}=0.125\space mol$
Step2: Calculate the molarity (M)
The volume of the solution $V = 96\space mL=0.096\space L$
Using the formula $M=\frac{n}{V}$, where $n = 0.125\space mol$ and $V = 0.096\space L$
$M=\frac{0.125\space mol}{0.096\space L}\approx1.3\space M$
Step3: Calculate the mass of water
Using the formula $m=
ho V$, where $
ho = 1.0\space g/mL$ (density of water) and $V = 95\space mL$
$m_{water}=1.0\space g/mL\times95\space mL = 95\space g$
Step4: Calculate the total mass of the solution
$m_{solution}=m_{NaOH}+m_{water}$
$m_{solution}=5.0\space g + 95\space g=100\space g$
Step5: Calculate the mass percentage
Using the formula $\text{Mass}\%=\frac{m_{solute}}{m_{solution}}\times100\%$
$\text{Mass}\%=\frac{5.0\space g}{100\space g}\times 100\% = 5.0\%$
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The molarity of the solution is approximately $1.3\space M$ and the mass percentage is $5.0\%$