Question

$the \: tree \: lover \: lesson \: summery \: by \: ruskin \: bond \: .$
​

Answer 1

Explanation:

In this book Rusty tells the story of his grandfather's relationship with trees. Of how his grandfather made him plant cuttings and saplings in the forest, saying that they were planting for the jungle and for the birds and animals who lived in it and needed more food and shelter.

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.