Question

how many oxygen atoms are present in 90.1 g of glucose (c₆h₁₂o₆)?
6.02 x 10²³
2.00 x 10²³
2.01 x 10²³
18.1 x 10²³
question 16
what is the molarity (m) of a 750 ml solution of 10.0 g of silver nitrate in water?
0.0079 m
0.79 m
7.9 m
0.079 m

Explanation:

Step1: Calculate moles of glucose

The molar mass of glucose ($C_6H_{12}O_6$) is $6\times12.01 + 12\times1.01+6\times16.00= 180.18\ g/mol$. The number of moles of glucose, $n=\frac{m}{M}$, where $m = 90.1\ g$ and $M = 180.18\ g/mol$. So $n=\frac{90.1}{180.18}\approx0.5\ mol$.

Step2: Determine moles of oxygen atoms

In one - mole of $C_6H_{12}O_6$, there are 6 moles of oxygen atoms. So in 0.5 mol of $C_6H_{12}O_6$, the number of moles of oxygen atoms is $n_{O}=0.5\times6 = 3\ mol$.

Step3: Calculate number of oxygen atoms

Using Avogadro's number ($N_A = 6.02\times10^{23}\ atoms/mol$), the number of oxygen atoms $N=n_{O}\times N_A=3\times6.02\times10^{23}= 1.806\times10^{24}\approx18.1\times10^{23}$ atoms.

Step4: Calculate moles of silver nitrate

The molar mass of silver nitrate ($AgNO_3$) is $107.87+14.01 + 3\times16.00=169.88\ g/mol$. The number of moles of $AgNO_3$, $n=\frac{m}{M}$, where $m = 10.0\ g$ and $M = 169.88\ g/mol$. So $n=\frac{10.0}{169.88}\approx0.0589\ mol$.

Step5: Calculate molarity of silver nitrate solution

The volume of the solution $V = 750\ mL=0.750\ L$. Molarity $M=\frac{n}{V}$, so $M=\frac{0.0589}{0.750}\approx0.079\ M$.

Answer:

For the first question: $18.1\times10^{23}$
For the second question: $0.079\ M$