JEE Main Some Basic Concepts of Chemistry Practice Questions With Solutions

Combustion of glucose $({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6)$\left(\right. C_{6} H_{12} O_{6} \left.\right) produces ${\mathrm{CO}}_2$CO_{2} and water. The amount of oxygen (in $\mathrm{g}$g) required for the complete combustion of $900\mathrm{g}$900 \textrm{ } g of glucose is :

[Molar mass of glucose in $\mathrm{g}{\mathrm{mol}}^{-1}=180$g \left(\textrm{ } mol\right)^{- 1} = 180]

Molality $(\mathrm{m})$\left(\right. m \left.\right) of $3\mathrm{M}$3 M aqueous solution of $\mathrm{NaCl}$NaCl is :(Given : Density of solution $=1.25\mathrm{g}{\mathrm{mL}}^{-1}$= 1.25 \textrm{ } g \left(\textrm{ } mL\right)^{- 1}, Molar mass in $\mathrm{g}{\mathrm{mol}}^{-1}:\mathrm{Na}-23,\mathrm{Cl}-35.5$g \left(\textrm{ } mol\right)^{- 1} : Na - 23 , Cl - 35.5)

The density of '$x$x' $\mathrm{M}$M solution ('$x$x' molar) of $\mathrm{NaOH}$NaOH is $1.12\mathrm{g}{\mathrm{mL}}^{-1}$1.12 \textrm{ } g \left(\textrm{ } mL\right)^{- 1}, while in molality, the concentration of the solution is $3\mathrm{m}$3 \textrm{ } m ( 3 molal). Then $x$x is

(Given : Molar mass of $\mathrm{NaOH}$NaOH is $40\mathrm{g}/\mathrm{mol}$40 \textrm{ } g / mol)

The number of moles of methane required to produce $11\mathrm{g}{\mathrm{CO}}_2(\mathrm{g})$11 \textrm{ } g \left(\textrm{ } CO\right)_{2} \left(\right. g \left.\right) after complete combustion is : (Given molar mass of methane in $\mathrm{g}{\mathrm{mol}}^{-1}:16$g \left(\textrm{ } mol\right)^{- 1} : 16 )

An organic compound has $42.1\mathrm{\%}$42.1 \% carbon, $6.4\mathrm{\%}$6.4 \% hydrogen and remainder is oxygen. If its molecular weight is 342 , then its molecular formula is :

The Molarity (M) of an aqueous solution containing $5.85\mathrm{g}$5.85 \textrm{ } g of $\mathrm{NaCl}$NaCl in $500\mathrm{mL}$500 \textrm{ } mL water is : (Given : Molar Mass $\mathrm{Na}:23$Na : 23 and $\mathrm{Cl}:35.5{\mathrm{gmol}}^{-1}$Cl : 35.5 \left(\textrm{ } gmol\right)^{- 1})

A sample of ${\mathrm{CaCO}}_3$CaCO_{3} and ${\mathrm{MgCO}}_3$MgCO_{3} weighed $2.21\mathrm{g}$2.21 \textrm{ } g is ignited to constant weight of $1.152\mathrm{g}$1.152 \textrm{ } g. The composition of mixture is :

(Given molar mass in $\mathrm{g}{\mathrm{mol}}^{-1}{\mathrm{CaCO}}_3:100,{\mathrm{MgCO}}_3:84$g \left(\textrm{ } mol\right)^{- 1} CaCO_{3} : 100 , MgCO_{3} : 84)

If a substance '$A$A' dissolves in solution of a mixture of '$B$B' and '$C$C' with their respective number of moles as ${\mathrm{n}}_{\mathrm{A}},{\mathrm{n}}_{\mathrm{B}}$n_{A} , n_{B} and ${\mathrm{n}}_{{\mathrm{C}}_3}$n_{C_{3}}. Mole fraction of $\mathrm{C}$C in the solution is

The quantity which changes with temperature is :

A metal chloride contains $55.0\mathrm{\%}$55.0 \% of chlorine by weight . $100\mathrm{mL}$100 \textrm{ } mL vapours of the metal chloride at STP weigh $0.57\mathrm{g}$0.57 \textrm{ } g. The molecular formula of the metal chloride is

(Given: Atomic mass of chlorine is $35.5\mathrm{u}$35.5 u)