The purpose of this lab was to identify the amount of salt needed in the water to “flink”, the point at which an object sinks or floats, the golf ball in a salt solution. Flinking occurs when the object is neither sinking or floating, it appears as if it is suspended in the middle of the solution. To flink an object, the object’s density has to have the same as the density of the solution that it is placed into. To achieve the same density of both the salt solution and golf ball, manipulation of the density of the salt solution had to be done. “When salt is added and dissolved in water, it breaks down into ions that are then attracted to the water molecules. This attraction causes them to bind tightly, increasing the amount of matter per volume”
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First, the density of the golf ball was determined. The mass was taken by utilizing an analytical balance with a result of 45.189 grams. To determine the volume of the golf ball, a piece of string was wrapped around the circumference of the ball. Then, the string was measured. The circumference of the ball was measured to be 13.75cm. To find the radius of the ball, the circumference was substituted into the formula C=2흿r. The radius of the ball was 2.188 cm. To find the volume of the ball, the equation V=흿(4/3)r3 was used. The volume of the ball was determined as 43.89 cm3, which is the same as 43.89mL3. The density of the ball, which is mass per volume was 45.189g / 43.89 mL = 1.030 g/mL. Then, the density was plugged in for “Y” into the equation to determine the amount of salt needed to produce a solution that has the same density as the golf ball. The equation was solved for X, and the result was 2.340 grams of salt needed to produce a salt solution that has the same density of the golf ball. A piece of weigh paper was placed on the analytical balance and the balance was then zeroed. 2.340 grams of salt was then measured out onto the weighing paper. Then, 50 mL of deionized water was measured with a graduated cylinder, and was added to a beaker, followed by the salt. The solution was stirred by using a stirring rod. The salt water was then carefully poured into the volumetric flask. …show more content…
Secondly, the density of solutions that required 15.000 grams of salt were 1.0913 g/mL and 1.0934 g/mL. Third, the density of the solution that required 5.000 grams of salt was 1.05 g/mL. The density of the solution that needed 10.000 g of salt was 1.06 g/mL. Lastly, the solution that required 22.000 grams of salt produced a density of 1.13 g/mL. The final density of the golf ball resulted as 1.030 g/mL. Our calculated solution that required 2.340 grams of salt had failed to flink the golf ball, because the ball had immediately sunk. Throughout the lab, a trend was noticed that as salt was added to the solution, density
First, the density of the golf ball was determined. The mass was taken by utilizing an analytical balance with a result of 45.189 grams. To determine the volume of the golf ball, a piece of string was wrapped around the circumference of the ball. Then, the string was measured. The circumference of the ball was measured to be 13.75cm. To find the radius of the ball, the circumference was substituted into the formula C=2흿r. The radius of the ball was 2.188 cm. To find the volume of the ball, the equation V=흿(4/3)r3 was used. The volume of the ball was determined as 43.89 cm3, which is the same as 43.89mL3. The density of the ball, which is mass per volume was 45.189g / 43.89 mL = 1.030 g/mL. Then, the density was plugged in for “Y” into the equation to determine the amount of salt needed to produce a solution that has the same density as the golf ball. The equation was solved for X, and the result was 2.340 grams of salt needed to produce a salt solution that has the same density of the golf ball. A piece of weigh paper was placed on the analytical balance and the balance was then zeroed. 2.340 grams of salt was then measured out onto the weighing paper. Then, 50 mL of deionized water was measured with a graduated cylinder, and was added to a beaker, followed by the salt. The solution was stirred by using a stirring rod. The salt water was then carefully poured into the volumetric flask. …show more content…
Secondly, the density of solutions that required 15.000 grams of salt were 1.0913 g/mL and 1.0934 g/mL. Third, the density of the solution that required 5.000 grams of salt was 1.05 g/mL. The density of the solution that needed 10.000 g of salt was 1.06 g/mL. Lastly, the solution that required 22.000 grams of salt produced a density of 1.13 g/mL. The final density of the golf ball resulted as 1.030 g/mL. Our calculated solution that required 2.340 grams of salt had failed to flink the golf ball, because the ball had immediately sunk. Throughout the lab, a trend was noticed that as salt was added to the solution, density