Question

question 3
what is the mass of 1.80 mol of al(no₃)₃?
272 g
383 g
257 g
161 g
question 4
how many total atoms are present in 1.00 mol of al₂(so₄)₃?
1.02×10²⁵ atoms
6.02×10²³ atoms
9.03×10²⁴ atoms
1.20×10²⁴ atoms

Explanation:

Question 3

Step1: Calculate molar mass of Al(NO₃)₃

The molar mass of Al is 26.98 g/mol, N is 14.01 g/mol, and O is 16.00 g/mol. For Al(NO₃)₃, molar - mass $M=(26.98)+3\times(14.01 + 3\times16.00)=26.98+3\times(14.01 + 48.00)=26.98+3\times62.01=26.98 + 186.03=213.01$ g/mol.

Step2: Calculate mass of 1.80 mol of Al(NO₃)₃

We use the formula $m = n\times M$, where $n = 1.80$ mol and $M = 213.01$ g/mol. So $m=1.80\times213.01 = 383.418\approx383$ g.

Step1: Determine the number of atoms in one formula - unit of Al₂(SO₄)₃

In Al₂(SO₄)₃, there are 2 Al atoms, 3 S atoms, and $3\times4 = 12$ O atoms. So in one formula - unit, there are $2 + 3+12=17$ atoms.

Step2: Calculate the number of atoms in 1.00 mol of Al₂(SO₄)₃

We know that 1 mol of any substance contains $N_A = 6.02\times10^{23}$ formula - units. For 1.00 mol of Al₂(SO₄)₃, the number of atoms $N=17\times N_A=17\times6.02\times10^{23}=1.0234\times10^{25}\approx1.02\times10^{25}$ atoms.

Answer:

383 g

Question 4