Molar mass of Na2CO3 = (2 × 23) + 12.00 + (3 × 16) = 106 g mol-1
Now, 1 mole of Na2CO3 means 106 g of Na2CO3.
∴ 0.5 mol of Na2CO3 = \(\frac {106\ g}{1 \ mole} × 0.5\ mol \) Na2CO3
= 53 g Na2CO3
⇒ 0.50 M of Na2CO3 = 0.50 mol/L Na2CO3
Hence, 0.50 mol of Na2CO3 is present in 1L of water or 53 g of Na2CO3 is present in 1L of water.
LIST I | LIST II | ||
A. | \(K_2SO_4(aq)\) with 60% dissociation | I. | \(i =3.7\) |
B. | \(K_3[Fe(CN)_6](aq)\) with 90% dissociation | II. | \(i= 1.8\) |
C. | \(AlCl_3 (aq)\) with 80% dissociation | III. | \(i=2.2\) |
D. | \(K_2HgI_4 (aq)\) with 40% dissociation | IV. | \(i= 3.4\) |
What is the percentage of carbon in ethanol (\(C_{2}H_{5}OH\)) ?
(Atomic masses: C = 12, H = 1, O = 16)
Give reasons for the following.
(i) King Tut’s body has been subjected to repeated scrutiny.
(ii) Howard Carter’s investigation was resented.
(iii) Carter had to chisel away the solidified resins to raise the king’s remains.
(iv) Tut’s body was buried along with gilded treasures.
(v) The boy king changed his name from Tutankhaten to Tutankhamun.
Find the mean deviation about the median for the data
xi | 15 | 21 | 27 | 30 | 35 |
fi | 3 | 5 | 6 | 7 | 8 |