90 grams of glucose, $c_6h_{12}o_6$, is dissolved in 2 liters of solution. determine the molarity (m).

Explanation:

Step1: Calculate molar mass of glucose

The molar mass of $C_6H_{12}O_6$: $C$: $6\times12.01\ g/mol = 72.06\ g/mol$, $H$: $12\times1.01\ g/mol=12.12\ g/mol$, $O$: $6\times16.00\ g/mol = 96.00\ g/mol$. Molar mass $M_{C_6H_{12}O_6}=72.06 + 12.12+96.00=180.18\ g/mol$.

Step2: Calculate moles of glucose

Use the formula $n=\frac{m}{M}$, where $m = 90\ g$ and $M = 180.18\ g/mol$. $n=\frac{90\ g}{180.18\ g/mol}\approx0.5\ mol$.

Step3: Calculate molarity

Use the formula $M=\frac{n}{V}$, where $n$ is the number of moles and $V$ is the volume of the solution in liters. Here $n = 0.5\ mol$ and $V = 2\ L$. $M=\frac{0.5\ mol}{2\ L}=0.25\ M$.

Answer:

$0.25$