9.3 g of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is _______ g. (nearest integer)

Question

9.3 g of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is _______ g. (nearest integer) [2024]

Smart Achievers · Accepted Answer

(20)

$n_{aniline} = \frac{Given mass}{Molar mass} = \frac{9.3 g}{93 g m o l^{- 1}} = 0.1 m o l$

$n_{dye} = n_{aniline} = 0.1 m o l$

$W_{dye} = n_{dye} \times M_{dye} = 0.1 m o l \times 198 g m o l^{- 1} = 19.8 g \approx 20 g$

Solution

Q.
9.3 g of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is _______ g. (nearest integer) [2024]

Ans.
(20)

$n_{aniline} = \frac{Given mass}{Molar mass} = \frac{9.3 g}{93 g m o l^{- 1}} = 0.1 m o l$

$n_{dye} = n_{aniline} = 0.1 m o l$

$W_{dye} = n_{dye} \times M_{dye} = 0.1 m o l \times 198 g m o l^{- 1} = 19.8 g \approx 20 g$

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Solution

Q. 9.3 g of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is _______ g. (nearest integer) [2024]

Ans. (20) naniline=Given massMolar mass=9.3 g93 gmol-1=0.1 mol ndye=naniline=0.1 mol Wdye=ndye×Mdye=0.1 mol×198 gmol-1=19.8 g≈20 g

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Q.
9.3 g of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is _______ g. (nearest integer) [2024]

Ans.
(20)

$n_{aniline} = \frac{Given mass}{Molar mass} = \frac{9.3 g}{93 g m o l^{- 1}} = 0.1 m o l$

$n_{dye} = n_{aniline} = 0.1 m o l$

$W_{dye} = n_{dye} \times M_{dye} = 0.1 m o l \times 198 g m o l^{- 1} = 19.8 g \approx 20 g$