SOLUTION 

You must make the following conversions:

Volume of dilute acid → moles of H?PO? → mass of H?PO? → mass of 85 % H?PO? 

0.6000 L dil. acid × $$\displaystyle\frac{{\text{2.80 mol H}_{{3}}\text{PO}_{{4}}}}{\text{1 L dil. acid}}\times\frac{{\text{98.00 g H}_{{3}}\text{PO}_{{4}}}}{{\text{1 mol H}_{{3}}\text{PO}_{{4}}}}$$ 

$$\displaystyle\frac{{\text{100 g 85 \% H}_{{3}}\text{PO}_{{4}}}}{{\text{85 g H}_{{3}}\text{PO}_{{4}}}}=\text{194 g of 85 \% H}_{{3}}\text{PO}_{{4}}$$

SOLUTION

(I) contains alkyne as well as carboxyl group, the alkyne having sp hybridized C- the centre has high e− density which stabilizes the −COO⁻ group. Hence H+is easily releasable.

(II) contains alkene?sp² hybridized centre?slightly less electronegative than (I), hence less acidic compared to (I)

(III) has an only carboxyl group. Does not have any alkyne or alkene to stabilize the COO⁻ group. Hence least acidic.

 The hexagonal close packing and the face-centred cubic system

have the closest packing as their packing efficiency is 74% and

thus (3) represent the correct solution.

Solution

Galena (PbS) are sulphide ores. Cassiterite SnO₂ (oxide ore), and Magnetite Fe₃O₄ (oxide ore) are not sulphide ore. The froth flotation process is used to concentrate sulphide ores, based on preferential wetting properties with frosting agent and water.

SOLUTION

As the phenoxide ion (C₆H₅−O⁻) is a better nucleophile than phenol (C₆H₅−OH), it reacts with CH₃I in presence of NaOH to form methoxybenzene. 

Hence both Assertion and Reason are correct and Reason is the correct explanation for Assertion

 AB  ? A⁺ + B⁻  Ksp=S² 

S=√Ksp =√4×10⁻¹⁰

Solution

            $$3BaCl_2+2(NH_4)_3PO_4\, \, \, \, \, \longrightarrow \, \, \, Ba_3(PO_4)_2+6NH_4Cl$$
             $$3 \,mol\, \, \, \,    \quad 2\, mol\,\,\, \quad \quad \quad \quad \quad \quad \quad                          1 \,mol$$
Given: $$0.5\,mol   \, \,\, \quad  0.2\,mol        \, \, \quad \quad \quad \quad\quad \quad                0.1\,mol$$
Thus, $$(NH_4)_3PO_4$$ is the limiting reactant giving $$0.1$$ mol $$Ba_3(PO_4)_2$$.$$BaCl_2$$ required by $$0.2$$ mol of $$(NH_4)_3PO_4 = 0.3 \: mol$$.
Thus, $$BaCl_2$$ in excess $$= 0.5 -0.3 = 0.2 \: mol$$