Question

calculate the number of molecules present in
a)15gram NaOH

b)18gram C6 H12 O6 (GLUCOSE)

c)2.5gram CaCo3​

Answer 1

molecular formula= C16,H12,O6

molar mass= 6×12+12×1+16×6=180gmol-1

Explanation:

moles in 18gram of sample=18/180=0.1mole

so, mollis mollis of carbon present in in sample = 0.1×6=0.6

moles of hydrogen=0.1×12=1.2

moles of oxygen=6×0.1=0.6

number of atmos present ine carbon=0.6×6.022×10 power of 23=3.6×10 power of 23

number of atoms present in hydrogen=1.2×6.022×10 power of 23=7.2×10 power of 23

number of atoms present in oxygen=0.6×6.022×10 power of 23=3.6×10 power of 23

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Answer 2

Answer:

a) 0.37 mols

b)0.1 mols

c) 0.025mols

Explanation:

a) Molar mass of NaOH is 23+16+1= 40 g/ mol.

no. of moles (NaOH) = 15g/40g/mol= 0.37 mol

b) Molar mass of C6H12O6 is 6×12+12×1+6×16= 180 g/ mol

No. of moles ( C6 H12 O6) = 18 gm/ 180gm/ mol= 0.1mols

C) Molar mass of CaCO3= 40+12+16×3= 100 mols

No. of moles = 2.5gm/100gm/mol= 0.025 mol

Hope this will help you