Question

a solution containing 80.0 grams of glucose (c₆h₁₂o₆) is dissolved in enough water to make 250 ml of solution. what is its molarity?
1.73 x 10⁻⁵ m
0.00178 m
1.78 m
0.111 m
question 18
1 pts
how many grams of kmno₄ are needed to make 500.0 ml of a 0.200 m solution?
395 g
15.8 g
63.2 g
1580 g

Explanation:

Question 1:

Step1: Calculate moles of glucose

The molar mass of glucose ($C_6H_{12}O_6$) is $6\times12 + 12\times1+6\times16= 180\ g/mol$. The number of moles $n$ of glucose is calculated by $n=\frac{m}{M}$, where $m = 80.0\ g$ and $M = 180\ g/mol$. So $n=\frac{80.0}{180}\ mol\approx0.444\ mol$.

Step2: Calculate molarity

The volume $V$ of the solution is $V = 250\ mL=0.250\ L$. Molarity $M$ is defined as $M=\frac{n}{V}$. Substituting $n = 0.444\ mol$ and $V=0.250\ L$ into the formula, we get $M=\frac{0.444}{0.250}=1.78\ M$.

Step1: Calculate moles of $KMnO_4$

The volume $V$ of the solution is $V = 500.0\ mL = 0.5000\ L$ and the molarity $M = 0.200\ M$. Using the formula $n = M\times V$, we have $n=0.200\ mol/L\times0.5000\ L = 0.100\ mol$.

Step2: Calculate mass of $KMnO_4$

The molar mass of $KMnO_4$ is $39 + 55+4\times16=158\ g/mol$. Using the formula $m=n\times M$, we get $m = 0.100\ mol\times158\ g/mol=15.8\ g$.

Answer:

C. 1.78 M

Question 2: