An intravenous solution of mannitol is used as a diuretic to increase the loss of sodium and chloride by a patient. If a patient receives 25.0 mL of a 25.% (m/v) mannitol solution, how many grams of mannitol were given?
Express your answer to two significant figures and include the appropriate units.
To find the grams of mannitol given, we can use the equation:
grams of mannitol = volume of solution (mL) × concentration of solution (m/v)
Plugging in the given values:
grams of mannitol = 25.0 mL × 25.% (m/v)
First, we need to convert the percent concentration to a decimal:
25.% (m/v) = 0.25 (m/v)
Plugging this in:
grams of mannitol = 25.0 mL × 0.25 (m/v)
grams of mannitol = 6.25 grams
Therefore, 6.25 grams of mannitol were given.
To calculate the number of grams of mannitol given, we need to multiply the volume of the solution by the concentration of mannitol.
Given:
Volume of solution = 25.0 mL
Concentration of mannitol = 25.% (m/v)
To express the concentration as a decimal, we divide it by 100: 25.%/100 = 0.25.
Now, we can calculate the number of grams of mannitol given:
Grams of mannitol = Volume of solution (in mL) × Concentration of mannitol (as a decimal)
= 25.0 mL × 0.25
Calculating this, we find:
Grams of mannitol = 6.25 g
Therefore, 6.25 grams of mannitol were given in the IV solution.
To find the number of grams of mannitol given, we can use the equation:
mass = volume × concentration
In this case, the volume of the mannitol solution given is 25.0 mL, and the concentration is 25% (m/v).
Step 1: Convert the volume from mL to L:
Since 1 L = 1000 mL, we divide the volume by 1000:
25.0 mL ÷ 1000 = 0.025 L
Step 2: Calculate the mass of mannitol using the equation:
mass = 0.025 L × 25% (m/v)
Step 3: Convert the concentration from a percentage to a decimal:
25% = 25/100 = 0.25
Step 4: Substitute the values into the equation:
mass = 0.025 L × 0.25
Step 5: Calculate the mass:
mass = 0.00625 grams
Rounding the answer to two significant figures, we get:
mass = 0.0063 grams (rounded)
Therefore, approximately 0.0063 grams of mannitol were given to the patient.