Question

1. calculate the molar concentration of glucose in the stock solution.

c₆h₁₂o₆
m=(6×12.01 + 12×1.01+6×16.00)g/mol = 180.18g/mol

Explanation:

Step1: Recall molar - concentration formula

Molar concentration ($c$) is given by $c=\frac{n}{V}$, where $n$ is the number of moles and $V$ is the volume of the solution in liters. First, we need to find the number of moles of glucose. The molar mass of glucose ($C_6H_{12}O_6$) is calculated as $M=(6\times12.01 + 12\times1.01+6\times16.00)\text{ g/mol}=180.18\text{ g/mol}$. But we are not given the mass of glucose or the volume of the solution. Assuming we have a mass $m$ of glucose and volume $V$ of the solution:

Step2: Calculate number of moles

The number of moles of glucose $n=\frac{m}{M}$, where $m$ is the mass of glucose in grams and $M = 180.18\text{ g/mol}$.

Step3: Calculate molar - concentration

Then $c=\frac{n}{V}=\frac{m}{M\times V}$. Since no values for $m$ and $V$ are given in the problem statement, if we assume we have $m$ grams of glucose in $V$ liters of solution, the molar concentration $c$ of glucose in the stock solution is $c=\frac{m}{180.18V}\text{ mol/L}$.

Answer:

If the mass of glucose is $m$ grams and the volume of the solution is $V$ liters, the molar concentration of glucose in the stock solution is $\frac{m}{180.18V}\text{ mol/L}$