Advertisements
Advertisements
प्रश्न
How will you convert the following in not more than two steps:
Acetophenone to Benzoic acid
Advertisements
उत्तर
Acetophenone to Benzoic acid
APPEARS IN
संबंधित प्रश्न
Derive van’t Hoff general solution equation.
Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
How will you convert the following in not more than two steps:
Benzoic acid to Benzaldehyde
Give reasons for the following
Elevation of the boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.
Van’t Hoff factor i is given by the expression:
(i) i = `"Normal molar mass"/"Abnormal molar mass"`
(ii) i = `"Abnormal molar mass"/"Normal molar mass"`
(iii) i = `"Observed colligative property"/"Calculated colligative property"`
(iv) i = `"Calculated colligative property"/"Observed colligative property"`
Van't Hoff factor I is given by expression.
Consider the reaction
\[\begin{bmatrix}\begin{array}{cc}
\phantom{.......}\ce{CH3}\\
\phantom{....}|\\
\ce{CH3CH2CH2 - \overset{⊕}{N} - CH2CH3}\\
\phantom{....}|\\
\phantom{.......}\ce{CH3}
\end{array}\end{bmatrix}\]\[\ce{OH^- ->[Heat] ?}\]
Which of the following is formed in a major amount?
When 19.5 g of F – CH2 – COOH (Molar mass = 78 g mol−1), is dissolved in 500 g of water, the depression in freezing point is observed to be 1°C. Calculate the degree of dissociation of F – CH2 – COOH.
[Given: Kf for water = 1.86 K kg mol−1]
Why is boiling point of 1 M NaCl solution more than that of 1 M glucose solution?
