You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Salt of Strong Acid and Strong Bases

Find the weig...

Question

Find the weight of $N a N O_{3}$ in grams (Rounded off to nearest integer), required to make 50 mL of an aqueous solution containing 70 mg per mL of $N a^{+}$ ions

[Given:Atomic weight in $g m o l^{- 1} [N a : 23; N : 14; O : 16]$

Open in App

Solution

$Concentration of N a^{+} = 70 m g / m L$
$W_{N a}$ in 50 mL solution:
$70 \times 50 m g = 3500 m g = 3.5 g$
$Moles of N a^{+} in 50 mL solution = \frac{3.5}{23} m o l$
$N a N O_{3} (a q) \to N a^{+} (a q) + N O_{3}^{-} (a q)$
$Moles of N a N O_{3} = Moles of N a^{+} = \frac{3.5}{23} m o l$
$Mass of N a N O_{3} = \frac{3.5}{23} \times 85 = 12.94 \approx 13 g$

Suggest Corrections

Similar questions

Q. The

N a N O_{3}

weighed to make

50 m L

of an aqueous solution containing 70.0

m g

N a^{+}

per

m L

is _____

g

. (Rounded off to the nearest integer)

[Given: Atomic weight in

g m o l^{- 1}

. Na: 23; N: 14; O: 16.]

Q. How much

N a N O_{3}

must be weighed out to make

100 m L

of an aqueous solution containing

23 m g o f N a^{+} p e r m L

500

mL of

0.25 M N a_{2} S O_{4}

solution is added to an aqueous solution of

15.0

g of

B a C l_{2}

solution resulting in the formation of white precipitate of

B a S O_{4}

. The moles of

B a S O_{4}

formed is

y \times 10^{- 3}

moles. Then,

y

is (in nearest integer) : [Given that molecular weight of

B a = 137, S = 32, O = 16, N a = 23

g/mol]

Q. A solution contains

4

g of

{N a}_{2} C O_{3}

and

N a C l

25

mL of this solution requires

50

mL of

\frac{N}{10}

H C l

for complete neutralization. The

%

composition of mixture is (as nearest integer) :

250 m l

of a sodium carbonate solution contains

2.65

grams of

{N a}_{2} {C O}_{3}

10 m l

of this solution is added to

^{'} x^{'} m l

of water to obtain

0.001 M

{N a}_{2} {C O}_{3}

solution.

What is the value of

^{'} x^{'}

m l

[Molecular weight of

{N a}_{2} {C O}_{3} = 106]

Join BYJU'S Learning Program

Find the weight of NaNO3 in grams (Rounded off to nearest integer), required to make 50 mL of an aqueous solution containing 70 mg per mL of Na+ ions [Given:Atomic weight in gmol−1[Na:23;N:14;O:16]

Concentration of Na+=70 mg/mL WNa in 50 mL solution: 70×50mg=3500 mg=3.5g Moles of Na+ in 50 mL solution =3.523 mol NaNO3 (aq)→Na+ (aq)+NO−3 (aq) Moles of NaNO3=Moles of Na+=3.523 mol Mass of NaNO3=3.523×85=12.94≈13g

Find the weight of $N a N O_{3}$ in grams (Rounded off to nearest integer), required to make 50 mL of an aqueous solution containing 70 mg per mL of $N a^{+}$ ions

[Given:Atomic weight in $g m o l^{- 1} [N a : 23; N : 14; O : 16]$