Question

your
sto
Or. Amon
Dr. Amount
(5000
Z. Re
harchases TR
55000
11000
2100
Dr.
-1300
7900
P.
To o
"
Pul
stock CN-94
Particulars
Capital of Jone John BSL
XDrawings as y
Baule Overdraft &
Furnefure BS
Creditoss BSL
Premises Bs
-10000
BS A
Book Debts BS A
- L9000
Rent Dell
SReturns TR
+lood
Sales TR
Discount PEL
Insurance Peel
Sundry expenses PELD
Commission
Carriage Outward Peel Dr3300
Cash BSA
+5600
To G
500
75000
20
Soo
asoo
to I
Less
looo
PEL C
Sur
Can
t
1,01,700
1,01,700​

Answer 1

Explanation:

Hmmm…hmmmm….

Hmmm…hmmmm….

Aye ho meri zindagi mein

Tum bahar banke

Aye ho meri zindagi mein

Tum bahar banke

Aye ho meri zindagi mein

Tum bahar banke

Mere dil mein yuhi rahena hai

Mere dil mein yuhi rahena

Tum pyar pyar banke

Aye ho meri zindagi mein

Tum bahar banke

Ankho mein tum base ho

Sapne hazar banke

Ankho mein tum base ho

Sapne hazar banke

Mere dil mein yuhi rahena hai

Mere dil mein yuhi rahena

Tum pyar pyar banke

Aye ho meri zindagi mein

Tum bahar banke

Mere sathi mere sajaan

Mere saath yuhi chalna

Mere sathi mere sajaan

Mere saath yuhi chalna

Badle ga raang zamana

Par tum nahi badalna

Meri maang yuhi bharna

Taare hazar ban ke

Meri maang yuhi bharna

Taare hazar ban ke

Mere dil mein yuhi rahena hai

Mere dil mein yuhi rahena

Tum pyar pyar banke

Aye ho meri zindagi mein

Tum bahar banke

Mere dil mein yuhi rahena

Tum pyar pyar banke

Agar me jo rooth jao to

Tum mujhe manana

Thama hai haath mera

Phir umar bhar nibhana

Mujhe chod ke na jana

Waade hazar karke

Mujhe chod ke na jana

Waade hazar karke

Mere dil mein yuhi rahena hai

Mere dil mein yuhi rahena

Tum pyar pyar banke

Aye ho meri zindagi mein

Tum bahar banke

Aye ho meri zindagi mein

Tum bahar banke

Mere dil mein yuhi rahena hai

Mere dil mein yuhi rahena

Tum pyar pyar banke

Aye ho meri zindagi mein

Tum bahar banke.

Answer 2

Explanation:

$\Large{\underline{\bf{\color{cyan}GiVeN,}}}$

Mass of dissolving haemoglobin is 10 g.

$\longmapsto\:\:\bf\orange{Mass\:(m)\:=\:10\:g}$

Volume of water is 1 dm³.

$\longmapsto\:\:\bf\green{Volume\:(V)\:=\:1\:dm^3\:=\:1\:L}$

Molar mass of haemoglobin is 6.51 × 10⁴ g/mol.

$\longmapsto\:\:\bf\purple{Molar\:mass\:(M)\:=\:6.51\times{10^4}\:g/mol}$

$\Large{\underline{\bf{\color{coral}To\:FiNd,}}}$

Molarity of the solution.

$\Large{\underline{\bf{\color{lime}CaLcUlAtIoN,}}}$

$\bf\pink{We\:know\;that,}$

➣ No. of moles of any substance present in a sample is the division of given mass of the substance to it's molar mass.

$\red\bigstar\:\:{\underline{\green{\boxed{\bf{\purple{No.\:of\:moles\:(n)\:=\:\dfrac{Mass\:(m)}{Molar\:mass\:(M)}\:}}}}}}$

$:\implies\:\:\bf{n\:=\:\dfrac{10\:g}{6.51\times{10^4}\:g/mol}\:}$

$:\implies\:\:\bf{n\:=\:\dfrac{1}{6.51\times{10^3}}\:mol}$

$:\implies\:\:\bf\orange{n\:=\:0.153\times{10^{-3}}\:mol}$

$\bf\red{Again,}$

➣ Molarity is defined as the number of moles of solute present in one litre of solution.

➣ Molarity is denoted by M.

$\pink\bigstar\:\:{\underline{\purple{\boxed{\bf{\blue{M\:=\:\dfrac{No.\:of\:moles\:of\:solute\:(n)}{Volume\:of\:solution\:in\:litres\:(V)}\:}}}}}}$

$:\implies\:\:\bf{M\:=\:\dfrac{0.153\times{10^{-3}}\:mol}{1\:L}\:}$

$:\implies\:\:\bf\green{M\:=\:0.153\times{10^{-3}}\:mol/L\:}$

━─━─━─━─━─━─━─━─━─━─━─━─━─━─━

$\Large\bf\purple{Therefore,}$

➛ Molarity of the solution is 0.153 × 10⁻³ mol/L.