Question

question 11
the mass percent (w/w%) of a solutions containing 3.5 g mgoh and 126 g h₂o is blank 1%
blank 1 add your answer
question 12
blank 1 milliliters are present in a 2.5 m solution containing 15 g of nacl.
blank 1 add your answer

Explanation:

Question 11

Step1: Calculate total mass of solution

The total mass of the solution is the sum of the mass of the solute ($Mg(OH)_2$) and the mass of the solvent ($H_2O$). So, $m_{total}=m_{Mg(OH)_2}+m_{H_2O}=3.5\ g + 126\ g=129.5\ g$.

Step2: Calculate mass - percent

The mass - percent formula is $\text{Mass}\%=\frac{m_{solute}}{m_{total}}\times100\%$. Substituting the values, we get $\text{Mass}\%=\frac{3.5\ g}{129.5\ g}\times100\%\approx2.7\%$.

Step1: Calculate moles of NaCl

The molar mass of $NaCl$ is $M_{NaCl}=22.99\ g/mol + 35.45\ g/mol = 58.44\ g/mol$. The number of moles of $NaCl$, $n=\frac{m}{M}=\frac{15\ g}{58.44\ g/mol}\approx0.257\ mol$.

Step2: Use molarity formula to find volume

The molarity formula is $M=\frac{n}{V}$, where $M$ is molarity, $n$ is the number of moles, and $V$ is the volume in liters. Rearranging for $V$, we get $V=\frac{n}{M}$. Given $M = 2.5\ M$ and $n = 0.257\ mol$, then $V=\frac{0.257\ mol}{2.5\ mol/L}=0.1028\ L$.

Step3: Convert volume to milliliters

Since $1\ L = 1000\ mL$, then $V = 0.1028\ L\times1000\ mL/L = 102.8\ mL$.

Answer:

$2.7$

Question 12