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Answers
1) No. of moles = given mass of compound / Molar mass
Given mass = 10g
Molar mass = (40 + 12 + 3x16) =100g/mol
No of moles of CaCO3 = 10 / 100 =0.1 mol
According to Avogadro’s law –
1 mol of CaCO3 molecules contains 6.022 x 1023 CaCO3 molecules
Hence, 0.1 mol of CaCO3 molecules will contain 6.022 x 1023x 0.1 molecules
i.e. 0.6022 x 1023 or 6.022 x 1022 molecules
2) Let the no. of moles be X.
Molecular weight of ammonia =17g
17 g - 1 mole
34 g - X
X = 34/17=2 moles
Thus, the no. of moles on 34 g of ammonia is 2 moles.
3) Molarity = NO. OF MOLES OF SOLUTE / VOLUME OF SOLUTION IN LITRE
HERE
MOLARITY=0.5M
VOLUME of solution =100ml
Molecular mass of Na2C03=23*2+12+16*3=106
Using formula and substituting value we have
0.5= x*1000 / 106*100
X=0.5*106*100/1000
X=5.3g
Answer:
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