Question

prelab questions -14 (1 pt). determine the concentration (in molarity) of sodium phosphate monobasic.
0.7 g of sodium phosphate monobasic
nah₂po₄·h₂o mw→138 g/mol
0.7 g nah₂po₄·h₂o $\frac{1 mol}{138 g nah₂po₄·h₂o}$ = 0.005 m
what equation should you be using anytime you do a dilution problem in lab?
m₁v₁ = m₂v₂
prelab question -15 (1 pt). determine how much of the stock concentration (see above for concentration) of rennin you need to make 25 ml of the following concentrations. show your work using proper lab math.
5%
25%

Explanation:

Step1: Calculate moles of sodium phosphate monobasic

The molar - mass of $NaH_2PO_4\cdot H_2O$ is $138\ g/mol$. The number of moles $n$ is calculated using the formula $n=\frac{m}{M}$, where $m = 0.7\ g$ and $M = 138\ g/mol$. So, $n=\frac{0.7\ g}{138\ g/mol}\approx0.00507\ mol$. However, the calculation in the picture for molarity is incorrect as it is missing the volume information. Assuming the volume $V = 1\ L$ (since molarity $M=\frac{n}{V}$ and if no volume is given and we want molarity in standard units), the molarity $M=\frac{0.7\ g/138\ g/mol}{1\ L}\approx0.0051\ M$.

Step2: For dilution problem

The dilution formula is $M_1V_1 = M_2V_2$, where $M_1$ and $V_1$ are the molarity and volume of the stock solution, and $M_2$ and $V_2$ are the molarity and volume of the diluted solution.

Step3: Solve for volume of stock solution for 5% rennin

Let the stock concentration be $C_1$ (not given in the problem - assume it is 100% for simplicity of percentage - based dilution). $V_2 = 25\ ml$ and $C_2=5\%$. Using the dilution formula $C_1V_1 = C_2V_2$, we have $100\% \times V_1=5\%\times25\ ml$. So, $V_1=\frac{5\%\times25\ ml}{100\%}=1.25\ ml$.

Step4: Solve for volume of stock solution for 25% rennin

Using $C_1V_1 = C_2V_2$ with $C_1 = 100\%$, $V_2 = 25\ ml$ and $C_2 = 25\%$. Then $V_1=\frac{25\%\times25\ ml}{100\%}=6.25\ ml$.

Answer:

The molarity of sodium phosphate monobasic (assuming volume of 1 L) is approximately $0.0051\ M$. The volume of stock rennin solution for 5% concentration to make 25 ml is $1.25\ ml$, and for 25% concentration is $6.25\ ml$.