Question

Please ANSWER BOTH QUESTIONS

1)Brontis is working in his lab. To continue his experiment, he needs to calculate the total number of ions in 4.85 mol of MgCl2. Calculate the total number of ions in 4.85 mol of MgCl2.

2)A chemist requires 0.885 mol Na2CO3 for a reaction. How many grams does this correspond to?

Answer 1

1)

number of molecules of MgCl2 = number of mol * Avogadro’s number

= 4.85 * 6.022*10^23 molecules

= 2.921*10^24 molecules

1 molecule of MgCl2 has 3 ions

So,

mol of ions = 3*number of molecules of MgCl2

= 3*2.921*10^24 ions

= 8.76*10^24 ions

Answer: 8.76*10^24 ions

2)

Molar mass of Na2CO3,

MM = 2*MM(Na) + 1*MM(C) + 3*MM(O)

= 2*22.99 + 1*12.01 + 3*16.0

= 105.99 g/mol

mass of Na2CO3,

m = number of mol * molar mass

= 0.885 mol * 105.99 g/mol

= 93.8 g

Answer: 93.8 g